1. ECE 874: Physical Electronics
Prof. Virginia Ayres
Electrical & Computer Engineering
Michigan State University

2. Lecture 25, 26 Nov 12 Chp. 05: Recombination-Generation Processes
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3. VM Ayres, ECE874, F12

4. pn junction in Si at equilibrium ( no bias)
Recombination-generation (R-G) via a trap (a local defect): why this is important:
pn junction in Si at equilibrium ( no bias) - +
p = p+ = 1019 cm-3
n = 1015 cm-3 Si W
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5. Same pn junction in Si in reverse bias: - 5V
- Vrev + - - - - + + + +
p+ = 1019 cm-3
n = 1015 cm-3 Si W
Reverse bias goal: turn the device OFF: no current flowing.
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6. You didn’t turn your device OFF as well as you thought you did by four orders of magnitude.
What happened: trap-mediated recombination-generation (R-G) processes act to restore what ever the previous steady state was. The trap released more carriers to try to restore the previous neutral region concentrations.
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7. Two key relationships/equations:
Note that n1 and p1 look just like ordinary dopant concentrations n and p except that they originate at a trap energy level ET’
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8. Two key relationships/equations in HW05:
Prs. 5.3, 5.4, 5.5
Prs. 5.2, 5.3, 5.4, 5.5: All
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9. Pr. 5.4: ET’ = Ei is worst case scenario: the one to check
Prs. 5.2, 5.3, 5.4, 5.5: All
Pr. 5.2: ET’ = Ei is given
Pr. 5.3: ET’ near Ei is given
Pr. 5.4: ET’ = Ei is worst case scenario: the one to check
Pr. 5.5: ET’ = Ei is given
Obviously there is something important about this condition
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10. VM Ayres, ECE874, F12

11. VM Ayres, ECE874, F12

12. VM Ayres, ECE874, F12

13. 3b)
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14. Points: p. 149:
para-
live NT
n or p
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15. Pr. 5.2
Pr. 5.3, 5.4, 5.5
3b)
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16. Pr. 5.2: (a) Given a condition on the capture cross sections cp and cn
(c) Given a condition on the emission cross sections ep and en
But: question is about a concentration nT not an R-G rate R
Saw nT in Lecture 24 in derivation of rates rN and rP under equilibrium conditions. nT is concentration of electrons in traps, pT is concentration of holes. Total concentration of filled traps NT = nT + pT.
Note: in Lecture 24, we worked through an equilibrium derivation but a steady state example. Time to connect these two ideas.
Prs. 5.2, 5.3, 5.4, 5.5 are all steady state problems.
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17. Lecture 24: General info:
Processes the change the e- headcount
Processes the change the hole headcount
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18. Lecture 24: Each one of these processes happens with better or worse efficiencies:
Hole capture
Hole emission
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19. Lecture 24: General info:
Processes the change the e- headcount
Processes the change the hole headcount
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20. Lecture 24: Equilibrium:
0 =
Pierret says same things: the definition for equilibrium is that the rates for changing electrons and holes concentrations via a trap individually sum to 0.
0a)
0b)
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21. Lecture 24: Equilibrium:
Under equilibrium conditions you can solve for the emission coefficients in terms of the capture coefficients (p. 145). Then, assuming that even away from equilibrium, the capture/emission coefficient values don’t change too much:
rN and rP are not 0, so this is not equilibrium
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22. The definition of the steady state condition is: when rN = rP
Under equilibrium conditions you can solve for the emission coefficients in terms of the capture coefficients (p. 145). Then, assuming that even away from equilibrium, the capture/emission coefficient values don’t change too much:
The definition of the steady state condition is:
when rN = rP
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23. Equilibrium/Steady State:
Under equilibrium conditions you can solve for the emission coefficients in terms of the capture coefficients (p. 145). Then, assume that even away from equilibrium, the capture/emission coefficient values don’t change too much.
.13a)
.13b)
These connectors between emission and capture coefficients should be used in Pr. 5.2 (c).
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24. Use NT = nT + pT to eliminate pT Solve for nT
Steady State:
Set rN = rP
Use NT = nT + pT to eliminate pT
Solve for nT
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25. nT for steady state conditions:
.21)
Use this definition for nT under steady state conditions to answer Pr. 5.2.
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26. R for steady state conditions:
Can use the steady state nT definition and pT = NT - nT to eliminate nT and pT in the above. The result is:
(5.24)
Use this R in Prs. 5.3, 5.4 and 5.5
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